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\title{\heiti\zihao{2} 复变函数-第3章}
\author{20373963-樊若宸}
\date{\today}

\begin{document}
\maketitle
\section{计算积分.}
 (2)$\int_{C}|z| \mathrm{d} z$, 积分路径 $C$ 是:(a) 连接点 $-1$ 到 1 的直线段; (b) 连接点 $-1$ 到 1 , 圆心在原点的上半个圆周.

\textbf{解:}\quad

(a)
$$
	\begin{aligned}
		\int_{C} f(z) \mathrm{d} z & =\int_{C} u(x, y) \mathrm{d} x-v(x, y) \mathrm{d} y+\mathrm{i} \int_{C} v(x, y) \mathrm{d} x+u(x, y) \mathrm{d} y \\
		                           & =\int_{C} \sqrt{x^2+y^2}\mathrm{d}x+i\int_{C}\sqrt{x^2+y^2}\mathrm{d}y                                            \\
		                           & =\int_{-1}^1 x\mathrm{d}x\qquad(y=0,\mathrm{d}y=0)                                                                \\
		                           & =1
	\end{aligned}
$$

(b)
$$
	\begin{aligned}
		\int_{C} f(z) \mathrm{d} z & =\int_{C} u(x, y) \mathrm{d} x-v(x, y) \mathrm{d} y+\mathrm{i} \int_{C} v(x, y) \mathrm{d} x+u(x, y) \mathrm{d} y \\
		                           & =\int_{C}\sqrt{x^2+y^2}\mathrm{d}x+i\int_{C}\sqrt{x^2+y^2}\mathrm{d}y\\
                                   &=\int_{-\pi}^{0}\sqrt{\cos^2\theta+\sin^2\theta}-\sin\theta\mathrm{d}\theta+i\int_{-\pi}^{0}\sqrt{\cos^2\theta+\sin^2\theta}\cos\theta\mathrm{d}\theta\\
                                   &=2
	\end{aligned}
$$
(4)$\int_{C} \dfrac{\mathrm{d} z}{z}$, 积分路径 $C$ 是: (a) 连接点 $-\mathrm{i}$ 到 $\mathrm{i}$,中心在原点的右半个圆周; (b) 连 接点 $-1$ 到 $1$ , 中心在原点的下半个圆周.

\textbf{解:}\quad

(a)
$$
\begin{aligned}
    \int_{C} f(z) \mathrm{d} z & =\int_{C} u(x, y) \mathrm{d} x-v(x, y) \mathrm{d} y+\mathrm{i} \int_{C} v(x, y) \mathrm{d} x+u(x, y) \mathrm{d} y\\
    &=\int_{C}\dfrac{x}{x^2+y^2}\mathrm{d}x+\dfrac{y}{x^2+y^2}\mathrm{d}y+i\int_{C}\dfrac{-y}{x^2+y^2}\mathrm{d}x+\dfrac{x}{x^2+y^2}\mathrm{d}y\\
    &=\int_{-\pi/2}^{\pi/2}-\cos\theta\sin\theta\mathrm{d}\theta+\sin\theta\cos\theta\mathrm{d}\theta+i\int_{-\pi/2}^{\pi/2}\sin^2\theta\mathrm{d}\theta+\cos^2\theta\mathrm{d}\theta\\
    &=i\pi
\end{aligned}
$$

(b)
$$
\begin{aligned}
    \int_{C} f(z) \mathrm{d} z & =\int_{C} u(x, y) \mathrm{d} x-v(x, y) \mathrm{d} y+\mathrm{i} \int_{C} v(x, y) \mathrm{d} x+u(x, y) \mathrm{d} y\\
    &=\int_{C}\dfrac{x}{x^2+y^2}\mathrm{d}x+\dfrac{y}{x^2+y^2}\mathrm{d}y+i\int_{C}\dfrac{-y}{x^2+y^2}\mathrm{d}x+\dfrac{x}{x^2+y^2}\mathrm{d}y\\
    &=\int_{\pi}^{2\pi}-\cos\theta\sin\theta\mathrm{d}\theta+\sin\theta\cos\theta\mathrm{d}\theta+i\int_{\pi}^{2\pi}\sin^2\theta\mathrm{d}\theta+\cos^2\theta\mathrm{d}\theta\\
    &=i\pi
\end{aligned}
$$

\section{利用积分估值证明:$$
\left|\int_{C}\left(x^{2}+i y^{2}\right) \mathrm{d} z\right| \leqslant \pi
$$其中$C$是连接$-i$到$i$的右半个圆周.}
\begin{proof}
    $$
    \begin{aligned}
            \int_{C}\left(x^{2}+i y^{2}\right) \mathrm{d} z&=\int_Cx^2\mathrm{d}x+y^2\mathrm{d}y+i\int_Cx^2\mathrm{d}y+y^2\mathrm{d}x\\
            &=\int_{-\pi/2}^{\pi/2}-\sin\theta\cos^2\theta\mathrm{d}\theta+\sin^2\theta\cos\theta\mathrm{d}\theta+i\int_{-\pi/2}^{\pi/2}\cos^3\theta\mathrm{d}\theta-\sin^3\theta\mathrm{d}\theta\\
            &=\dfrac{2}{3}+i\dfrac{4}{3}
    \end{aligned}
    $$

    从而
    $$
    \left|\int_{C}\left(x^{2}+i y^{2}\right) \mathrm{d} z\right| =\sqrt{\dfrac{4+16}{9}}=\dfrac{2\sqrt{5}}{3}<\pi
    $$

    或者利用$|x^2+iy^2|=\sqrt{x^4+y^4}\leqslant 1$,则
    $$
    \left|\int_{C}\left(x^{2}+i y^{2}\right) \mathrm{d} z\right| \leqslant 1\cdot\pi=\pi
    $$
\end{proof}



\section{计算下列积分,其中$C$为正向圆周.}

(2)$\int_{C} \dfrac{\cos z}{z+\mathrm{i}} \mathrm{d} z$, 其中 $C:|z+3 \mathrm{i}|=1$

\textbf{解:}\quad
函数仅在$z_0=-i$点处不解析,在$\Omega = |z+3i|\leqslant 1$中解析,从而有
$$
\int_{C} \dfrac{\cos z}{z+\mathrm{i}} \mathrm{d} z=0
$$

(4)$\int_{C} \dfrac{2 z-1}{z(z-1)} \mathrm{d} z$, 其中 $C:|z|=2$.

\textbf{解:}\quad
先代数变形,再由Cauchy积分公式:
$$
\begin{aligned}
    \int_{C} \dfrac{2 z-1}{z(z-1)} \mathrm{d} z&=\int_C\dfrac{1}{z}+\dfrac{1}{z-1}\mathrm{d}z\\
    &=\int_C\dfrac{1}{z}\mathrm{d}z+\int_C\dfrac{1}{z-1}\mathrm{d}z\\
    &=2\pi i+2\pi i\\
    &=4\pi i
\end{aligned}
$$

\section{计算积分$\int_C(3z^2+2z+1)\mathrm{d}z$,其中$C$是从点$-i$到$i$的右半单位圆周.}
\textbf{解:}\quad
显然$3z^2+2z+1$在全平面解析,从而积分只与初末位置有关.
$$
\begin{aligned}
    \int_C(3z^2+2z+1)\mathrm{d}z&=\left.z^3+z^2+z\right|_{-i}^{i}\\
    &=0 
\end{aligned}
$$



\section{设$f(z)=\int\limits_{C}\dfrac{3\xi^2+7\xi+1}{\xi -z}\mathrm{d}\xi$,其中$C:|\xi|=3$,求$f'(1+i)$.}
\textbf{解:}\quad 
由于$z_0=1+i$点处不解析,从而由Cauchy积分公式可得:
$$
\begin{aligned}
    f(z)&=2\pi i(3z^2+7z+1)\\
\end{aligned}
$$
从而$f'(z)=2\pi i(6z+7)$,$f'(1+i)=2\pi i(13+6i)=-12\pi+26\pi i$

\section{求积分$\int_C\dfrac{\mathrm{e}^{z}}{z}\mathrm{d}z$,其中$C:|z|=1$为正向圆周,进而证明$$\int_0^{\pi}\mathrm{e}^{\cos\theta}\cos(\sin\theta)\mathrm{d}\theta=\pi$$}
\begin{proof}
    由Cauchy积分公式:
$$
\begin{aligned}
    \int_C\dfrac{\mathrm{e}^{z}}{z}\mathrm{d}z&=2\pi i\\
    \int_C\dfrac{\mathrm{e}^{z}}{z}\mathrm{d}z&=\int_C\dfrac{\mathrm{e}^{\cos\theta+i\sin\theta}}{\cos\theta + i\sin\theta}\mathrm{d}(\cos\theta+i\sin\theta)\\
    &=\int_C\mathrm{e}^{\cos\theta+i\sin\theta}(\cos\theta-i\sin\theta)(-\sin\theta+i\cos\theta)\mathrm{d}\theta\\
    &=i\int_0^{2\pi}\mathrm{e}^{\cos\theta}(\cos(\sin\theta)+i\sin(\sin\theta))\mathrm{d}\theta\\
    &=i\int_0^{2\pi}\mathrm{e}^{\cos\theta}\cos(\sin\theta)\mathrm{d}\theta-\int_0^{2\pi}\mathrm{e}^{\cos\theta}\sin(\sin\theta)\mathrm{d}\theta\\
    &=2\pi i
\end{aligned}
$$
从而$\int_0^{2\pi}\mathrm{e}^{\cos\theta}\cos(\sin\theta)\mathrm{d}\theta=2\pi$.
又因为$\mathrm{e}^{\cos\theta}\cos(\sin\theta)$关于$\pi$对称,从而
$$
\begin{aligned}
    \int_0^{\pi}\mathrm{e}^{\cos\theta}\cos(\sin\theta)\mathrm{d}\theta&=\dfrac{1}{2}\int_0^{2\pi}\mathrm{e}^{\cos\theta}\cos(\sin\theta)\mathrm{d}\theta\\
    &=\pi
\end{aligned}
$$
\end{proof}
\section{如果在 $|z|<1$ 内 $f(z)$ 解析,并且
$$
|f(z)| \leqslant \dfrac{1}{1-|z|}
$$
证明
$$
\left|f^{(n)}(0)\right| \leqslant(n+1) !\left(1+\dfrac{1}{n}\right)^{n}<\mathrm{e}(n+1) !(n=1,2, \cdots)
$$}
\begin{proof}
   $$
   \begin{aligned} \left| f^{(n)}(0) \right|&=\left| \dfrac{n !}{2 \pi i} \int_{C} \dfrac{f(\xi)}{\xi^{n+1}} \mathrm{d}\xi\right| \\ & \leqslant \dfrac{n !}{2 \pi} \int_{C}\left|\dfrac{f(\xi)}{\xi^{n+1}}\right||\mathrm{d} \xi| \\ & \leqslant \dfrac{n !}{2 \pi} \int_{C} \dfrac{1}{\left|\xi^{n+1}\right||1-\xi|} |\mathrm{d}\xi| \\ &\leqslant\dfrac{n !}{|z|^{n}(1-|z|)} \\
    &\leqslant \dfrac{(n+1)!}{\left(\dfrac{n}{n+1}\right)^n}\\
    &=(n+1)!\left(1+\dfrac{1}{n}\right)^n\\
    &\leqslant \mathrm{e}(n+1)!
    \end{aligned}
   $$
\end{proof}
本题主要是运用了Cauchy不等式,这里的应用反映了Cauchy不等式对估计复函数的高阶导数的作用.在变形过程中使用均值不等式的取等条件以及无理数$\mathrm{e}$的定义.

\section{设 $f(z)$ 为整函数, 如果在整个复平面上 $|f(z)|\geqslant 1$, 则 $f(z)$ 必为常数.}
\begin{proof}
    由条件$\dfrac{1}{|f(z)|}\leqslant 1$,从而$\dfrac{1}{f(z)}$也是整函数,并且其有界$M=1$,从而其为常数.从而推知$f(z)$亦是常数.
\end{proof}

\section{验证下列各函数为调和函数,并根据已知条件求解析函数 $f(z)=u+\mathrm{i} v$.}
(1) $u=y^{3}-3 x^{2} y-2, f(\mathrm{i})=-1+\mathrm{i}$;
$$
\begin{aligned}
    \Delta u&=\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}\\
    &=-6y+6y=0  
\end{aligned}
$$
故调和.又
$$
\dfrac{\partial u}{\partial x}=-6xy\qquad \dfrac{\partial u}{\partial y}=3y^2-3x^2
$$
从而
$$
\dfrac{\partial v}{\partial y} = \dfrac{\partial u}{\partial x}=-6xy\qquad\dfrac{\partial v}{\partial x}=-\dfrac{\partial u}{\partial y}=3x^2-3y^2
$$
对上式左侧对$y$积分:
$$
v = -3xy^2+\varphi(x)
$$
对$x$求偏导:
$$
\dfrac{\partial v}{\partial x}=-3y^2+\varphi'(x)=3x^2-3y^2
$$
从而对$x$积分:
$$
\varphi(x) = x^3+C
$$
显然$v$调和.从而$f(z)=y^3-3x^2y-2+i(-3xy^2+x^3+C)$,将初值$f(i)=-1+i$代入得$C=1$,从而$f(z)=\left(\dfrac{z}{i}\right)^3+i-2$

(2) $u=\dfrac{y}{x^{2}+y^{2}}, f(2)=0$;

(3) $u=\mathrm{e}^{x}(x \cos y-y \sin y), f(0)=0$;
$$
\begin{aligned}
    \Delta u&=\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}\\
    &={\mathrm{e}}^x\,\left(x\,\cos\left(y\right)-y\,\sin\left(y\right)\right)+2\,{\mathrm{e}}^x\,\cos\left(y\right)    -{\mathrm{e}}^x\,\left(2\,\cos\left(y\right)+x\,\cos\left(y\right)-y\,\sin\left(y\right)\right)\\
    &=0
\end{aligned}
$$
从而$u$调和.又可由C-R方程:
\begin{equation}
    \begin{aligned}
    \dfrac{\partial v}{\partial y} = \dfrac{\partial u}{\partial x}&={\mathrm{e}}^x\,\left(x\,\cos\left(y\right)-y\,\sin\left(y\right)\right)+{\mathrm{e}}^x\,\cos\left(y\right)\\
    \dfrac{\partial v}{\partial x}=-\dfrac{\partial u}{\partial y}&={\mathrm{e}}^x\,\left(\sin\left(y\right)+y\,\cos\left(y\right)+x\,\sin\left(y\right)\right)
\end{aligned}
\end{equation}

对下式积分得:
$$
v = \mathrm{e}^x(\sin y+y\cos y+\sin y(x-1))+\varphi(y)
$$
上式对$y$求偏导:
$$
\begin{aligned}
    \dfrac{\partial v}{\partial y}&={\mathrm{e}}^x\,\left(2\,\cos\left(y\right)+\cos\left(y\right)\,\left(x-1\right)-y\,\sin\left(y\right)\right)+\varphi'(y)
\end{aligned}
$$
与(1)比较可得$\varphi'(y) = 0$,从而$\varphi =C$.有$v$调和.故
$$
f(z) = \mathrm{e}^x(x\cos y - y\sin y)  +i[\mathrm{e}^x(x\sin y+y\cos y)+C]
$$
将初值代入解得:
$$
C=0
$$
从而
$$
f(z)=\mathrm{e}^x(x\cos y - y\sin y)  +i[\mathrm{e}^x(x\sin y+y\cos y)]=z\mathrm{e}^z
$$
(4) $u=(x-y)\left(x^{2}+4 x y+y^{2}\right)$.


\end{document}